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[parent] circle has one center (Theorem)
Theorem   Every circle has exactly one center.

Before proving this, we state and prove a lemma.

Lemma   A center of a circle must lie in its interior.
Proof. Suppose not. Let a circle have a center $ O$ not lying in its interior. If $ O$ lies on the circle, then the circle is degenerate (just a point). Suppose that $ O$ lies in its exterior. Choose a point $ A$ on the circle such that a line that contains $ A$ and $ O$ passes through the interior of the circle (i.e., the line is not tangent to the circle). Such a line intersects the circle at another point $ B$. It cannot be the case that $ O$ is in between $ A$ and $ B$, lest $ O$ be in the interior of the circle. Without loss of generality, let $ B$ be between $ A$ and $ O$.

\begin{pspicture}(-3,-3)(5,3) \rput[a](0,3){.} \rput[a](0,-3){.} \pscircle(0,0){... ...put[r](-3.2,0){$A$} \rput[l](3.1,-0.3){$B$} \rput[l](4.2,0){$O$} \end{pspicture}

Then $ BO<AO$, a contradiction, since $ O$ is supposed to be the center of the circle. $ \qedsymbol$

Now to prove the theorem.

Proof. By definition, every circle has at least one center. Suppose that a circle has more than one center. Let $ O$ and $ O'$ be two distinct centers of this circle. By the previous lemma, $ O$ and $ O'$ must lie in the interior of the circle. Draw a chord of the circle which contains both $ O$ and $ O'$. Let $ A$ and $ B$ be the intersections of this chord with the circle such that $ O$ is between $ A$ and $ O'$. Since $ O$ and $ O'$ are in the interior of the circle, it must be the case that $ O'$ is between $ O$ and $ B$.

\begin{pspicture}(-4,-3)(4,3) \rput[b](0,-3){.} \rput[a](0,3){.} \pscircle(0,0){... ...a](-0.7,-0.3){$O$} \rput[a](0.7,-0.3){$O'$} \rput[l](3.2,0){$B$} \end{pspicture}

Note that we must have $ A \neq O$, $ A \neq O'$, $ B \neq O$, and $ B \neq O'$ as pictured. Otherwise, the circle is degenerate, yielding that $ O=O'$. Because of these four inequalities, we also have that $ AO>0$, $ AO'>0$, $ BO>0$, and $ BO'>0$.

Since $ O$ is a center of the circle, $ AO=BO$. Since $ O'$ is a center of the circle, $ AO'=BO'$. Thus, $ AO<AO'=BO'<BO$, a contradiction. It follows that a circle has exactly one center. $ \qedsymbol$

Generality

Looking at the proof of the main theorem and the lemma, we see that they hold in any geometry in which any two points lie on a common line and in which the notion of betweenness is well-defined. Examples of such geometries include Euclidean geometry, hyperbolic geometry, and neutral geometry. In fact, the uniqueness of center of a circle holds in any ordered geometry satisfying the congruence axioms (see here for a proof). By contrast, in spherical geometry, where the notion of betweenness is not well-defined and a pair of antipodal points determines an infinity of lines, we also have that every circle has exactly two centers which are antipodal points. However, in projective geometry, it is again valid because we identify antipodal points to construct the projective plane from the sphere.

More generally, in Riemannian spaces (and generalizations such as Finsler spaces), this theorem will hold provided that geodesics emanating at a point do not focus at some other point, although they can focus at the same point at which they started. This generalizes what we saw in the case of the sphere and the the projective plane. In both those cases, geodesics focussed but, in the former case, they focussed at the antipodal point but, in the latter case, they only focussed back at their starting point. While on spheres, circles have two centers, we can have Riemannian spaces in which geodesics refocus any number of times, even infinitely often, in which cases a circle could have any number of centers, even infinitely many of them. Because, by geodesic deviation, focussing requires positive curvature, we can assert that circles in spaces of non-positive curvature will have unique centers; for instance, this explains why the result holds in hyperbolic geometry (which has constant negative curvature).



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"circle has one center" is owned by Wkbj79. [ full author list (5) ]
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See Also: regular polygon and circles


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proof of uniqueness of center of a circle (Proof) by CWoo
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Cross-references: negative, curvature, positive, infinitely often, even, focus, geodesics, sphere, projective plane, projective geometry, infinity, antipodal points, spherical geometry, congruence axioms, ordered geometry, neutral geometry, hyperbolic geometry, Euclidean geometry, well-defined, lie on, geometry, inequalities, chord, contradiction, without loss of generality, intersects, contains, line, point, lies on, center, circle
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This is version 20 of circle has one center, born on 2007-06-13, modified 2007-06-24.
Object id is 9591, canonical name is CircleHasOneCenter.
Accessed 968 times total.

Classification:
AMS MSC51A05 (Geometry :: Linear incidence geometry :: General theory and projective geometries)
 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)
 51M10 (Geometry :: Real and complex geometry :: Hyperbolic and elliptic geometries and generalizations)

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Section on generalization by rspuzio on 2007-06-14 16:20:36
Along the lines of what we discussed last night, I moved the
comments about what spaces this theorem holds in to the end
in a section on generalization. The reason for this is not to
confuse beginners who may not even have heard of non-Euclidean
geometry whilst, at the same time. presenting more complete
information for the expert.
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a circle really *does* have one center......given certain constraints by Wkbj79 on 2007-06-13 19:15:30
I'll leave this guy world editable for a while in case I missed some minor detail.

I hope that this incident sparked by Mathprof's correction brings home the point that mathematicians can take very little (if anything at all) for granted.
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